Question: $\dfrac{ 9b + 9c }{ 5 } = \dfrac{ b - 10d }{ 4 }$ Solve for $b$.
Explanation: Multiply both sides by the left denominator. $\dfrac{ 9b + 9c }{ {5} } = \dfrac{ b - 10d }{ 4 }$ ${5} \cdot \dfrac{ 9b + 9c }{ {5} } = {5} \cdot \dfrac{ b - 10d }{ 4 }$ $9b + 9c = {5} \cdot \dfrac { b - 10d }{ 4 }$ Multiply both sides by the right denominator. $9b + 9c = 5 \cdot \dfrac{ b - 10d }{ {4} }$ ${4} \cdot \left( 9b + 9c \right) = {4} \cdot 5 \cdot \dfrac{ b - 10d }{ {4} }$ ${4} \cdot \left( 9b + 9c \right) = 5 \cdot \left( b - 10d \right)$ Distribute both sides ${4} \cdot \left( 9b + 9c \right) = {5} \cdot \left( b - 10d \right)$ ${36}b + {36}c = {5}b - {50}d$ Combine $b$ terms on the left. ${36b} + 36c = {5b} - 50d$ ${31b} + 36c = -50d$ Move the $c$ term to the right. $31b + {36c} = -50d$ $31b = -50d - {36c}$ Isolate $b$ by dividing both sides by its coefficient. ${31}b = -50d - 36c$ $b = \dfrac{ -50d - 36c }{ {31} }$